Mathematics • Year 10 • Unit 1 • Lesson 20

Tapered Solids in the Real World

Apply the pyramid, cone and sphere volume formulas to real Australian situations — conical paper cups, spherical scoops of ice cream on a cone, the Great Pyramid scaled to a model, and combined silo + cone roofs. Then explain the one-third rule in your own words.

Apply · Real-World Maths

1. Word problems

Each problem uses one or more of the tapered-solid formulas from Lesson 20: pyramid V = ⅓ × A_base × h, cone V = ⅓πr²h, sphere V = (4/3)πr³. Convert units carefully (1 cm³ = 1 mL, 1 m³ = 1 kL) and round only at the final step.

1.1 — Conical paper cup. A conical paper cup at a water cooler has radius 4 cm and perpendicular height 12 cm.

(a) Find the volume of the cup in exact form.
(b) Convert this to millilitres (1 d.p.).
(c) How many cups can be filled from a 1.5 L bottle? (Round down — partial cups don't count.) 3 marks

Stuck? V = ⅓π(4)²(12) = ⅓π(192) = 64π cm³. 1 L = 1000 mL.

1.2 — Scoop of ice cream on a cone. A waffle cone is a cone with radius 3 cm and perpendicular height 10 cm. A perfectly spherical scoop of ice cream of radius 3 cm sits on top.

(a) Find the volume of the empty cone.
(b) Find the volume of the spherical scoop.
(c) Find the total volume of the dessert (cone + sphere) in cm³ to 1 d.p. 3 marks

Stuck? Use the addition strategy from Lesson 19: V_total = ⅓π(3)²(10) + (4/3)π(3)³.

1.3 — Great Pyramid scale model. A school builds a 1:1000 scale wooden model of the Great Pyramid of Giza for a project. The Great Pyramid has a square base of side 230 m and perpendicular height 147 m. The scale model uses the same dimensions in millimetres (so base side 230 mm, height 147 mm).

(a) Convert the model's dimensions to cm.
(b) Find the volume of the model in cm³ to 1 d.p. 3 marks

Stuck? 230 mm = 23 cm, 147 mm = 14.7 cm. V_pyramid = ⅓ × 23² × 14.7.

1.4 — Composite silo with conical roof. An on-farm grain silo is a cylinder of radius 1.5 m and height 6 m, with a conical roof of the same radius (1.5 m) and perpendicular height 1 m on top.

(a) Find the volume of the cylindrical section in exact form.
(b) Find the volume of the conical roof in exact form.
(c) Find the total capacity in kilolitres (1 m³ = 1 kL) to 2 d.p. 4 marks

Stuck? V_cyl = π(1.5)²(6) = 13.5π. V_cone = ⅓π(1.5)²(1) = 0.75π. Add, then × 1 to get kL.

1.5 — Spherical fish-tank ornament. A glass marble dropped into an aquarium displaces water equal to its own volume. The marble is a sphere of diameter 1.6 cm.

(a) Find the volume of the marble in exact form.
(b) Convert this volume to mL to 2 d.p.
(c) If 25 identical marbles are dropped in, by how many mL does the water level rise (total displacement)? 3 marks

Stuck? r = 0.8 cm. V = (4/3)π(0.8)³. 1 cm³ = 1 mL.

2. Explain your thinking

This question is about communication, not just numbers. Use full sentences. 4 marks

2.1 The lesson states: "A pyramid (or cone) with the same base and height as a prism (or cylinder) has exactly one-third the volume." Using a concrete example you compute yourself (start with a cylinder of radius 6 cm and height 10 cm), explain (i) what the one-third rule says with your example's numbers, (ii) why the rule makes intuitive sense (think about how the cross-section changes from base to apex), and (iii) why a pyramid cannot have more than the prism's volume, no matter how tall.

Stuck? Cylinder V = π(6)²(10) = 360π. Cone with same r and h: V = ⅓ × 360π = 120π. Then a cone could only beat the cylinder if its height grew — but then we'd be comparing different shapes.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

1.1 — Conical cup (r = 4, h = 12)

(a) V = ⅓π(4)²(12) = ⅓π(192) = 64π cm³.
(b) V ≈ 201.1 mL.
(c) 1500 mL ÷ 201.1 ≈ 7.46 ⇒ 7 cups (round down).

1.2 — Cone + spherical scoop (r = 3, cone h = 10)

(a) V_cone = ⅓π(3)²(10) = ⅓π(90) = 30π cm³.
(b) V_sphere = (4/3)π(3)³ = (4/3)π(27) = 36π cm³.
(c) Total = 30π + 36π = 66π ≈ 207.3 cm³ (1 d.p.).

1.3 — Great Pyramid model (1:1000 scale)

(a) Base side 23 cm, height 14.7 cm.
(b) V = ⅓ × 23² × 14.7 = ⅓ × 529 × 14.7 = ⅓ × 7776.3 = 2592.1 cm³ (1 d.p.).

1.4 — Silo with conical roof (r = 1.5, cyl h = 6, cone h = 1)

(a) V_cyl = π(1.5)²(6) = π(2.25)(6) = 13.5π m³.
(b) V_cone = ⅓π(1.5)²(1) = ⅓π(2.25) = 0.75π m³.
(c) Total = 14.25π ≈ 44.77 m³ = 44.77 kL (2 d.p.).

1.5 — Spherical marble (d = 1.6 cm)

(a) r = 0.8 cm. V = (4/3)π(0.8)³ = (4/3)π(0.512) ≈ 2.048π/3 cm³ ≈ 0.6827π cm³ (or 2.048π/3).
(b) V ≈ 2.14 mL (2 d.p.).
(c) 25 marbles displace 25 × 2.14 ≈ 53.6 mL.

2.1 — Explain your thinking (sample response)

(i) For a cylinder of radius 6 cm and height 10 cm, V = π(6)²(10) = 360π cm³. A cone with the same r = 6 cm and same h = 10 cm has V = ⅓ × 360π = 120π cm³. The cone is exactly one-third of the cylinder.
(ii) The rule makes sense because a prism's cross-section stays the same all the way up, but a pyramid's (or cone's) cross-section shrinks to zero at the apex. On average, the pyramid only has one-third of the cross-sectional area of the matching prism, so it holds one-third of the volume.
(iii) A pyramid cannot have more than the prism's volume (with the same base and height) because at every height between 0 and h, the pyramid's cross-section is smaller than the prism's. You'd need a taller pyramid to match the prism — but then the comparison is no longer "same base and same height".

Marking: 1 for the concrete numerical example (360π vs 120π), 1 for the intuitive "cross-section shrinks" reason, 1 for stating that the pyramid is always strictly smaller for the same base and height, 1 for clear sentence structure.